# How do you long divide (9x^4-2-6x-x^2)/(3x-1)?

Jun 15, 2015

$\left(3 {x}^{3} + {x}^{2} - 2\right) - \frac{4}{3 x - 1}$

#### Explanation:

Let's put the polynomial in its usual order

$P \left(x\right) = 9 {x}^{4} - {x}^{2} - 6 x - 2$

First, divide $9 {x}^{4}$ by $3 x \implies 3 {x}^{3}$

Now subtract $3 {x}^{3} \left(3 x - 1\right) = 9 {x}^{4} - 3 {x}^{3}$ to $P \left(x\right)$, you obtain ${P}_{1} \left(x\right) = 3 {x}^{3} - {x}^{2} - 6 x - 2$

Now we procede the same way: divide $3 {x}^{3}$ by $3 x \implies {x}^{2}$

Now subtract ${x}^{2} \left(3 x - 1\right) = 3 {x}^{3} - {x}^{2}$ to ${P}_{1} \left(x\right)$, you obtain ${P}_{2} \left(x\right) = - 6 x - 2$

Now divide $- 6 x$ by $3 x \implies - 2$

Now subtract $- 2 \left(3 x - 1\right) = - 6 x + 2$ to $6 x - 2$, you obtain -4, which means

$P \left(x\right) = \left(3 x - 1\right) \left(3 {x}^{3} + {x}^{2} - 2\right) - 4$

Or, if you want to write it in fractions, $P \left(x\right) = \left(3 {x}^{3} + {x}^{2} - 2\right) - \frac{4}{3 x - 1}$

(NB: maybe there's an error: knowing this kind of exercises, I'm pretty sure they want the reminder to be 0, so maybe $P \left(x\right) = 9 {x}^{4} - {x}^{2} - 6 x + 2$)